$\overline{AB}$ = $6\sqrt{2}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $6\sqrt{2}$ $?$ $ \sin( \angle BAC ) = \frac{ \sqrt{2}}{2}, \cos( \angle BAC ) = \frac{ \sqrt{2}}{2}, \tan( \angle BAC ) = 1$
Answer: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{6\sqrt{2}} $ $ \overline{BC}=6\sqrt{2} \cdot \sin( \angle BAC ) = 6\sqrt{2} \cdot \frac{ \sqrt{2}}{2} = 6$